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 June 16th, 2010, 05:57 PM #1 Joined: Feb 2009 Posts: 76 Thanks: 0 The range of a matrix transformation How to do you determine if a vector (w) is in the range a matrix transformation. Let f: R^2 --> R^3 f(x) = Ax $\ A=\begin{bmatrix} 1 &2 \\ 0 &1 \\ 1& 1 \end{bmatrix}$/extract_itex] Vector w $\ w=\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}\$ Why is it a no? My work: ( I don't know if I'm doing it right) $\ f(x)=\begin{bmatrix} 1 &2 \\ 0 &1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}\$ $\ =\begin{bmatrix} 1 &2 \\ 0& 1 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}=\begin{bmatrix} 3\\ 1 \end{bmatrix}\$ and why is for Vector w, it's a yes? $\ w=\begin{bmatrix} 8\\ 5\\ 3 \end{bmatrix}\$ $\ =\begin{bmatrix} 1 &2 \\ 0& 1 \end{bmatrix}\begin{bmatrix} 8\\ 5 \end{bmatrix}=\begin{bmatrix} 18\\ 5 \end{bmatrix}\$ The book says that the set of all images of the vectors in R^n is called the range of f. I don't know what it means. Can someone please explain to me how to determine the range. Thanks in advance  June 16th, 2010, 07:21 PM #2 Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 1 Re: The range of a matrix transformation You want w= f(v) = Av, where v is a vector $\[\begin{array}x\\ y\end{array}$$. The equation should be: $$\begin{array}1\\ 1\\ 1\end{array}$ = $\begin{array} 1&2\\ 0&1\\ 1&1\end{array}$ $\begin{array}x\\ y\end{array}$$ You cannot substitute values of w into v, you need to see if any value of v, when multiplied by A gives you w. See if there is any v=[x,y] which satisfies this equation. Edit: Some clarification. "The set of all images" is the set $\{f(v)|v\in \mathbb{R}^2\}$. So, if w is in the image, w=f(v) for some vector v in R^2. To check if such a vector exists, see if there is any vector which satisfies f(v)=w.

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### determine if the given vector is in the range of t

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